Does the collection of coverings on a set $X$ form a lattice when ordered by refinement? This is a follow-up question to this question, prompted by a comment in Todd Trimble's answer.
Let $X\neq \emptyset$ be a set. We say that $U\subseteq {\cal P}(X)\setminus \{\emptyset\}$ is a proper covering if


*

*$\bigcup U = X$, and

*for $a\neq b\in U$ we have $a\not\subseteq b$.


Let $\text{Cov}(X)$ denote the collection of all (proper) coverings of $X$. For $A, B\in \text{Cov}(X)$ we set $A\leq B$ if $A$ refines $B$, that is for all $a\in A$ there is $b\in B$ such that $a\subseteq b$. (Note that if we consider all coverings instead of just the proper ones, we may lose anti-symmetry.)
Is $\text{Cov}(X)$ with the ordering defined above a lattice? Is it complete?
 A: If there are only finitely many points, it is a lattice, and it is always an upper semi-lattice. But in the infinite case, it is not a lattice.
François's comment on Fedor's answer shows that is it always at least an upper
semi-lattice, since the join of two covers $A$ and $B$ consists precisely of the inclusion maximal elements of $A\cup B$. It follows from this observation that there can be no finite counterexample to the original question. This is because there is always a least cover, the cover consisting of
the singletons, and any finite upper semi-lattice with a least
element is a lattice: the meet of any two objects is the
join of their (finitely many) lower bounds. So if there are only finitely many points, then indeed it is a lattice. 
Meanwhile, there are counterexamples in the case of infinitely many points. Consider the disjoint union of two copies of the
integers $\newcommand\Z{\mathbb{Z}}\Z\sqcup \Z$, red and blue, and
let $A$ consist of sets of the form $\{r\mid r<k\}\sqcup\{b\mid
b>k\}$, where $r$ is red and $b$ is blue and $k\in\Z$. And let $B$ consist of
the sets of the form $\{r\mid r>k\}\sqcup\{b\mid b<k\}$. So each of
these are proper covers of $\Z\sqcup\Z$.
I claim that there is no greatest lower bound of $A$ and $B$. If $C$ is a cover
that is a lower bound, then every member of $C$ will have to have all red elements below all blue elements (if any), because it refines $A$; and similarly, it will have to have all blue elements below all red elements (if any), because it refines $B$. So every element of $C$ will be monochromatic. Furthermore, the elements of $C$ must be bounded above in the red component and below in the blue component,
because it refines $A$, and bounded below in the red
component and above in the blue component, because it refines $B$. So every member of $C$ is a finite monochromatic set. Furthermore, it is easy to see that any proper cover consisting entirely
of finite monochromatic sets is a lower bound to both $A$ and $B$ under refinement.
Thus, there can be no greatest lower bound for $A$ and $B$, because for any such
lower bound, we can make a strictly coarser lower bound by simply
merging two sets of the same color (and omitting any subsets of the merger).
A similar example can be made with any infinite set, simply by finding two copies of $\Z$ inside the set. So on an infinite set, it is not a lattice.
A: (Completely wrong answer, saved for saving the comments.)
It is not a lattice in general. Consider four corners of the square and two coverings by parallel sides. They do not have unique supremum.
